1. Problem 6-62 A carton turnover machine is shown below. 16 in 15 in 7.3 in 7 in 60 0 2 in 3 in

2. Problem 6-62 At the instant shown, the 15 in link is driven clockwise at 5 rad/sec, and decelerating at 40 rad/sec 2. Determine the instantaneous torque required to operate the 15 in arm, and the forces at the two lower bearings. The two long links closely resemble slender members, made from steel, with a width of 1 in and a thickness of 0.25 in. The carrier and carton weigh 26 lb and its mass moment of inertia, relative to an axis through its center, is 2.75 lb in s 2.

3. Kinematic Diagram 2 A B C D G2 G4 G3 3 4 31 0 7.3” 16” 15” 5.83” 60 0 3.6 22 0 7 56.3 0 2 3

4. Velocity Analysis V B = 75 in/s V B =  2 r AB = (5 rad/s)(15 in) = 75 in/s V C/B VCVC V C = V B +> V C/B  to CB  to CD V G3/B V G3/C V G3  to G3C  to G3B V G3 = V B +> V G3/B = V C +> V G3/C

5. Velocity Results V C = 40.4 in/s 68 0 V C/B = 49.8 in/s V X/B = 25.6 in/s 56.3 0 V X/C = 41.5 in/s 31 0 V G3 = 77.7 in/s 49 0 V G2 = 37.5 in/s 30 0 V G4 = 20.2 in/s 68 0

6. Acceleration Analysis a n C = (40.4) 2 /16 = 102.0 in/s 2 22.0 0 a n G3/C = (41.5) 2 /5.83 = 295.4 in/s 2 59.0 0 a n G3/B = (25.6) 2 /3.6 = 182.0 in/s 2 33.7 0 a t B = (15)(40) = 600.0 in/s 2 30.0 0 a n C/B = (49.8) 2 /7 = 354.3 in/s 2 a n B = (75) 2 /15 = 375.0 in/s 2 60.0 0

7. Acceleration Analysis a n C +>a t C = a n B +>a t B +> a n C/B +>a t C/B  to CB  to CD anCanC anBanB atBatB a n C/B atCatC a t C/B aBaB aCaC

8. Acceleration Analysis a t C = 250.8 in/s 2 68.0 0   =(250.8)/16 =15.68 rad/s 2 ccw   =(332.7)/7 =47.53 rad/s 2 ccw a C = 270.7 in/s 2 a t C/B = 332.7 in/s 2 a G4 = 135.4 in/s 2 a B = 707.5 in/s 2 62 0 a G2 = 353.8 in/s 2 62 0

9. anCanC anBanB atBatB a n C/B atCatC a t C/B a n G3/B a n G3/C a t G3/B a t G3/C  to G3C  to G3B a G3 a G3 = 672 in/s 2 83 0 a G3 = a B +>a n G3/B +>a t G3/B = a C +> a n G3/C +>a t G3/C

10. Inertia Forces:Link 2 15 in 60 0 a G2 = 353.8 in/s 2 62 0 Vol 2 = (16 in)(1 in)(.25 in) = 4 in 3 W 2 = (.283 lb/in 3 )(4 in 3 ) = 1.13 lb F i G2 = (1.13 lb)(353.8 in/s 2 )/(386.4 in/s 2 ) = 1.03 lbs 62 0 F i G2

11. Inertia Forces:Link 2 15 in 60 0 F i G2 I G2 = ¼ [(1.13 lb)/(386.4 in/s 2 )](16 in) 2 = 0.19 lb in s 2   = 40 rad/s 2 ccw T i G2 = (0.19 lb in s 2 )(40 rad/s 2 ) = 7.5 lb incw T i G2 Assume slender rod

12. Inertia Forces:Link 3 a G3 = 672 in/s 2 83 0   =47.53 rad/s 2 ccw F i G3 = (26 lb)(672 in/s 2 )/(386.4 in/s 2 ) = 45.2 lbs 83 0 F i G3 T i G3 = (2.75 in lb s 2 )(47.53 rad/s 2 ) = 130.7 in lbscw T i G3

13. Vol 4 = (17 in)(1 in)(.25 in) = 4.25 in 3 W 4 = (.283 lb/in 3) (4.25 in 3 ) = 1.20 lb Inertia Forces:Link 4 16 in 22 0 a G4 = 135.4 in/s 2 F i G4 = (1.20 lb)(135.4 in/s 2 )/(386.4 in/s 2 ) = 0.42 lb F i G4

14. 16 in 22 0 Inertia Forces: Link 4 F i G4 I G4 = ¼ [(1.20 lb)/(386.4 in/s 2 )](17 in) 2 = 0.22 lb in s 2 T i G4 = (0.22 lb in s 2 )(15.68 rad/s 2 ) = 3.4 lb incw T i G4 Assume slender rod   = 15.68 rad/s 2 ccw

15. Free Body Diagrams Cx Cy Cx Dy Dx Ay Ax Bx By Bx F i G3 T i G3 F i G2 T i G4 F i G2 T i G2 W3W3 W4W4 W2W2 TATA

16. FBD: Link 2 7.5 60 0 Ay Ax By Bx F i G2 T i G2 62 0 15 in  Fx = 0 Ax - Bx + F i G2 cos62 = 0  Fy = 0 Ay - By +F i G2 sin62 - W 2 = 0 W2W2  A = 0 Bx[15(sin60)] + By[15(cos60)] + W 2 [7.5(cos60)] - T i G2 - T A - [F i G2 (sin62)][7.5(cos60)] – [F i G2 (cos62)][7.5(sin60)] = 0 TATA

17. FBD:Link 3 Bx Cx F i G3 3 in 7 in 2 in Cy By T i G3 W3W3 83 0  Fx = 0 Bx + Cx + F i G3 cos83 = 0  Fy = 0 By + Cy + F i G3 sin83 – W 3 = 0  c = 0 W 3 (3) - Bx(7) - (F i G3 sin83)(3) - (F i G3 cos83)(5) - T i G3 = 0

18. FBD:Link 4 16 in 22 0 Dy Dx Cy Cx F i G4 T i G4 W4W4 8  Fx = 0 Dx - Cx = 0  Fy = 0 Dy - Cy + F i G4 - W 4 = 0  D = 0 Cx[16(sin22)] + Cy[16(cos22)] + W 4 [8(cos22)] - F i G4 [8(cos22)] - T i G4 = 0

19. Finally, Solving the Nine Simultaneous Equations: Ax = -23.3 lbs Ay = -11.5 lbs Bx = -22.8 lbs By = -11.7 lbs Cx = 17.3 lbs Cy = -7.2 lbs Dx = 17.3 lbs Dy = -6.4 lbs T A = -394.1 in lbs